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1

JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Let A (sec$$\theta$$, 2tan$$\theta$$) and B (sec$$\phi$$, 2tan$$\phi$$), where $$\theta$$ + $$\phi$$ = $$\pi$$/2, be two points on the hyperbola 2x2 $$-$$ y2 = 2. If ($$\alpha$$, $$\beta$$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$$\beta$$)2 is equal to ____________.
Your Input ________

Answer

Correct Answer is Bonus

Explanation

Since, point A (sec$$\theta$$, 2tan$$\theta$$) lies on the hyperbola 2x2 $$-$$ y2 = 2

Therefore, 2sec2$$\theta$$ $$-$$ 4tan2$$\theta$$ = 2

$$\Rightarrow$$ 2 + 2tan2$$\theta$$ $$-$$ 4tan2$$\theta$$ = 2

$$\Rightarrow$$ tan$$\theta$$ = 0 $$\Rightarrow$$ $$\theta$$ = 0

Similarly, for point B, we will get $$\phi$$ = 0.

but according to question $$\theta$$ + $$\phi$$ = $${\pi \over 2}$$ which is not possible.

Hence, it must be a 'BONUS'.
2

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
If the minimum area of the triangle formed by a tangent to the ellipse $${{{x^2}} \over {{b^2}}} + {{{y^2}} \over {4{a^2}}} = 1$$ and the co-ordinate axis is kab, then k is equal to _______________.
Your Input ________

Answer

Correct Answer is 2

Explanation

Tangent

$${{x\cos \theta } \over b} + {{y\sin \theta } \over {2a}} = 1$$



So, area $$(\Delta OAB) = {1 \over 2} \times {b \over {\cos \theta }} \times {{2a} \over {\sin \theta }}$$

$$ = {{2ab} \over {\sin 2\theta }} \ge 2ab$$

$$\Rightarrow$$ k = 2
3

JEE Main 2021 (Online) 27th July Evening Shift

Numerical
Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, $$-$$4), one focus at (4, $$-$$4) and one vertex at (5, $$-$$4). If mx $$-$$ y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m2 is equal to _____________.
Your Input ________

Answer

Correct Answer is 3

Explanation

Given C(3, $$-$$4), S(4, $$-$$4)



and A(5, $$-$$4)

Hence, a = 2 & ae = 1

$$\Rightarrow$$ e = $${1 \over 2}$$

$$\Rightarrow$$ b2 = 3

So, $$E:{{{{(x - 3)}^2}} \over 4} + {{{{(y + 4)}^2}} \over 3} = 1$$

Intersecting with given tangent.

$${{{x^2} - 6x + 9} \over 4} + {{{m^2}{x^2}} \over 3} = 1$$

Now, D = 0 (as it is tngent)

So, 5m2 = 3.
4

JEE Main 2021 (Online) 20th July Evening Shift

Numerical
If the point on the curve y2 = 6x, nearest to the point $$\left( {3,{3 \over 2}} \right)$$ is ($$\alpha$$, $$\beta$$), then 2($$\alpha$$ + $$\beta$$) is equal to _____________.
Your Input ________

Answer

Correct Answer is 9

Explanation

Let, $$P \equiv \left( {{3 \over 2}{t^2},3t} \right)$$ is on the curve.

Normal at point P

$$tx + y = 3t + {3 \over 2}{t^3}$$

Passes through $$\left( {3,{3 \over 2}} \right)$$

$$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$$

$$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$$

$$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$$

$$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$$

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